Three point charges are placed at three corners of a rectangle of sides 30 cm and 40 cm. The fourth corner D is empty. A charge of \(+3\ \mathrm{nC}\) is 30 cm from D, a charge of \(-4\ \mathrm{nC}\) is 40 cm from D, and a charge of \(+10\ \mathrm{nC}\) is at the corner opposite to D. The electric potential at D is (take \(\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(180\ \mathrm{V}\)
The potential at D is the algebraic sum of the potentials of the three charges, each with its own sign. The \( +10\ \mathrm{nC} \) charge sits at the far end of the diagonal. The diagonal of a 30 cm by 40 cm rectangle is 50 cm.
So \( r_1=0.30\ \mathrm{m} \), \( r_2=0.40\ \mathrm{m} \) and \( r_3=0.50\ \mathrm{m} \).
\( \begin{aligned} V &= \frac{1}{4\pi\varepsilon_0}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}+\frac{q_3}{r_3}\right) \\ &= 9\times10^{9}\left(\frac{3\times10^{-9}}{0.30}-\frac{4\times10^{-9}}{0.40}+\frac{10\times10^{-9}}{0.50}\right) \\ &= 9\times10^{9}\times(1-1+2)\times10^{-8} \\ &= 180\ \mathrm{V} \end{aligned} \)
From the above, the potentials of the two nearer charges add up to zero. Hence, the whole potential at D comes from the \( +10\ \mathrm{nC} \) charge.
NCERT (2026-27):
Potential Due to a System of Charges – “… By the superposition principle, the potential V at P due to the total charge configuration is the algebraic sum of the potentials due to the individual charges”
\( V=V_1+V_2+\dots+V_n \qquad (2.17) \)